Projectile Motion: Cannonball Trajectory

Question

Trace the curve of the trajectory of a cannonball fired at an angle of \( 30^\circ \) to the horizontal. Neglect the effect of air resistance. Also, calculate the maximum height and horizontal distance (range) attained by the cannonball.

Solution

To calculate the trajectory, we will use the equations for projectile motion. The key variables are:

Launch angle \( \theta = 30^\circ \)
Initial velocity \( v_0 = 50 \, \text{m/s} \)
Acceleration due to gravity \( g = 9.81 \, \text{m/s}^2 \)

Step 1: Maximum Height

The formula for the maximum height attained by a projectile is:

\( H = \frac{v_0^2 \sin^2(\theta)}{2g} \)

Substituting \( \theta = 30^\circ \) and \( g = 9.81 \, \text{m/s}^2 \):

\( H = \frac{50^2 \times \sin^2(30^\circ)}{2 \times 9.81} = \frac{50^2 \times \left(\frac{1}{2}\right)^2}{2 \times 9.81} = \frac{50^2}{78.48} \)

Therefore, the maximum height is:

\( H = 31.86 \, \text{meters} \)

Step 2: Horizontal Range (Distance)

The formula for the range (horizontal distance) is:

\( R = \frac{v_0^2 \sin(2\theta)}{g} \)

Substituting \( \theta = 30^\circ \) and \( g = 9.81 \, \text{m/s}^2 \):

\( R = \frac{50^2 \times \sin(60^\circ)}{9.81} = \frac{50^2 \times \frac{\sqrt{3}}{2}}{9.81} = 220.57 \, \text{meters} \)

Step 3: Parametric Equations for the Trajectory

The horizontal and vertical positions as a function of time are given by the parametric equations:

\( x(t) = v_0 \cos(\theta) t \)